Extended Discrete Green's Theorem

This discrete Green's theorem (A Discrete Green's Theorem) connects a given function's double integral over a given domain and the linear combination of the values of the function's cumulative distribution function at the corners of the domain. This suggests a natural extension; by partitioning the domain into rectangles and a curvilinear part, we divide the calculation of the function's double integral over the domain into two parts: the integral over the rectangular domain is calculated using the discrete Green's theorem, and the curvilinear part is calculated via the usual double integral. The "sewing" between these two parts is performed using the parameter of tendency, as suggested in the slanted integration method (Slanted Line Integral). The formula stated by this theorem is simply: , where is the slanted line integral of over the edge of the domain .

In this Demonstration you can control the location of the points , the curvature of each edge connecting two adjacent vertices, the curve's orientation, and which of the subcurves is calculated. A green, red, or black vertex or edge indicates a positive, negative, or zero tendency. A blue vertex or edge means that the tendency is not calculated.

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Snapshot 1: The only vertices that do not occur in the final linear combination are (since the tendency is zero at these vertices). Let us detail the calculations that led to the theorem's statement:

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Here the first step is due to the additivity of the slanted integral, the second step is due to the definition of the slanted integral and the curve's tendencies at the specific points, and the last step is due to the discrete Green's theorem.

Snapshot 2: In case the edges are perpendicular to the axes, then the theorem agrees with the discrete Green's theorem. Note that here the edges and were not selected, thus the color of the vertices and is light red (to denote that their matching coefficient in the final linear combination is , and not {-, {1}}; and is colored blue to denote that no calculation at all is done at this vertex.

Snapshot 3: Note that here the vertices and meet. Thus, these vertices' coefficient is ( and cancel). A more delicate deduction occurs in the upper-left corner of the domain. To simplify the discussion, let us denote the intersection point of the edges and by . The double integral over the whole rectangle L′C1OC2 should be deducted: part of it does not intersect the given domain, and the double integral over the other part is calculated twice (once for each of the edges L′L and ML′—the dark yellow part in the graph). Indeed, we note that the integral is automatically deducted, since (according to the discrete Green's theorem): , and thus the deduction is bidirectional: the integral over the rectangle is deducted, the unwanted vertices C2,L′,C1 are deducted, and in return we get with a coefficient, as we would expect in the discrete Green's theorem for the polygon .

Snapshot 4: This snapshot depicts the following property of the slanted line integral: if is closed, then where is the curve taken with reversed orientation. It is easy to see from this snapshot that:

,

where stands for the boundary of , and is the outer polygon, .

A rigorous formulation of the theorem is as follows. Let be a simply connected domain in , whose boundary is a tendable curve (its tendency is defined everywhere). Let be an integrable function and let , , be a cumulative distribution function of f. Then , where is the slanted line integral of over the boundary of the domain D, regardless of the choice of points used to calculate the slanted line integral. This theorem can be improved by selecting the points on the curve such that the computational efficiency is maximized, rather than by selecting the points arbitrarily.